Question

aqueous hydrobromic acid (hbr) reacts with solid sodium hydroxide (naoh) to produce aqueous sodium bromide (nabr) and liquid water (h₂o). what is the theoretical yield of sodium bromide formed from the reaction of 3.2 g of hydrobromic acid and 2.2 g of sodium hydroxide? be sure your answer has the correct number of significant digits in it.

Explanation:

Step1: Write the balanced chemical equation

$$\text{HBr (aq)}+\text{NaOH (s)} ightarrow\text{NaBr (aq)}+\text{H}_2\text{O (l)}$$

Step2: Calculate the molar masses

• Molar mass of $\text{HBr}$: $M_{\text{HBr}} = 1 + 79.904=80.904\ \text{g/mol}$
• Molar mass of $\text{NaOH}$: $M_{\text{NaOH}} = 22.99 + 16 + 1=39.99\ \text{g/mol}$
• Molar mass of $\text{NaBr}$: $M_{\text{NaBr}} = 22.99 + 79.904 = 102.894\ \text{g/mol}$

Step3: Calculate the moles of reactants

• Moles of $\text{HBr}$: $n_{\text{HBr}}=\frac{m_{\text{HBr}}}{M_{\text{HBr}}}=\frac{3.2\ \text{g}}{80.904\ \text{g/mol}}\approx0.0396\ \text{mol}$
• Moles of $\text{NaOH}$: $n_{\text{NaOH}}=\frac{m_{\text{NaOH}}}{M_{\text{NaOH}}}=\frac{2.2\ \text{g}}{39.99\ \text{g/mol}}\approx0.0550\ \text{mol}$

Step4: Determine the limiting reactant

From the balanced equation, the mole ratio of $\text{HBr}$ to $\text{NaOH}$ is $1:1$. Since $n_{\text{HBr}}

Step5: Calculate the moles of $\text{NaBr}$ produced

Using the mole ratio from the balanced equation ($1\ \text{mol HBr}:1\ \text{mol NaBr}$), $n_{\text{NaBr}} = n_{\text{HBr}}=0.0396\ \text{mol}$

Step6: Calculate the mass of $\text{NaBr}$ (theoretical yield)

$m_{\text{NaBr}}=n_{\text{NaBr}}\times M_{\text{NaBr}}=0.0396\ \text{mol}\times102.894\ \text{g/mol}\approx4.1\ \text{g}$

Answer:

$4.1\ \text{g}$